Question
In the interval (1, 2), function f(x) = 2|x - 1| + 3|x - 2| is:
- Monotonically increasing.
- Monotonically decreasing.
- Not monotonic.
- Constant.
Solution:
f(x) = 2|x - 1| + 3|x - 2|
$\text{x}\in(1,2)$
x > 1 and x < 2
⇒ x - 1 > 0 and x - 2 < 0
⇒ f(x) = 2|x - 1| + 3|x - 2|
⇒ f(x) = 2(x - 1) - 3(x - 2)
⇒ f(x) = 2x - 2 - 3x + 6
⇒ f(x) = -x + 4
⇒ f'(x) = -1
Hence, function is monotonically decreasing.
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Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
| $x=-1$ | $x=0$ | $x=2$ | |
| $f(x)$ | $3$ | $6$ | $0$ |
| $g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$
$(A)$ $e-1$ $(B)$ $\int_1^e \ln (e+1-y) d y$ $(C)$ $e-\int_0^1 e^x d x$ $(D)$ $\int_1^r \ln y d y$